Let $P \equiv(-1,0), Q \equiv(0,0)$ and $R=(3,3)$ be three points. The equation of the bisector of the angle PQR is

• A

  $x-\sqrt{3} y=0$
• B

  $\sqrt{3} \mathrm{x}-\mathrm{y}=0$
• C

  $\mathrm{x}+\sqrt{3} \mathrm{y}=0$
• Correct answer
  D

  $\sqrt{3} x+y=0$

Solution

$\begin{array}{ll}\text { Given that, } & \mathrm{P}=(-1,0) \\ \text { and } & \mathrm{Q}=(0,0) \\ & \mathrm{R}=(3,3 \sqrt{3})\end{array}$

In $\triangle \mathrm{AQR}$, $$ \begin{aligned} \tan \phi &=\frac{3 \sqrt{3}}{3} \\ \Rightarrow \quad \tan \phi &=\sqrt{3} \end{aligned} $$ $$ \begin{aligned} \Rightarrow & & \tan \phi &=\tan 60^{\circ} \\ \Rightarrow & \phi &=\frac{\pi}{3} \end{aligned} $$ Slope of line QR $\mathrm{m}_{1}=\frac{3 \sqrt{3}-0}{3-0}=\sqrt{3}$ $\Rightarrow \quad \mathrm{m}_{1}=\sqrt{3}$ Slope of line $\mathrm{QP}$ $\mathrm{m}_{2}=\frac{0-0}{-1-0}=0 \Rightarrow \mathrm{m}_{2}=0$ Slope of line QP $$ \mathrm{m}_{2}=\frac{0-0}{-1-0}=0 \Rightarrow \mathrm{m}_{2}=0 $$ Let the angle between PQR is $\theta$, then $$ \begin{aligned} \tan \theta &=\left|\frac{\mathrm{m}_{1}-\mathrm{m}_{2}}{1+\mathrm{m}_{1} \cdot \mathrm{m}_{2}}\right| \\ \Rightarrow \quad \tan \theta &=\left|\frac{\sqrt{3}-0}{1+0}\right|=\sqrt{3} \text { or }-\sqrt{3} \\ \Rightarrow \quad & \tan \theta=\tan \frac{2 \pi}{3} \Rightarrow \theta=\frac{2 \pi}{3} \\ & {[\because(\tan \theta \neq \sqrt{3}) \text { from figure }] } \end{aligned} $$ Hence, $\quad \angle \mathrm{RQB}=\frac{\theta}{2}=\frac{2 \pi}{6}=\frac{\pi}{3}$ So, the angle bisector making an angle with (+) ive $x$-axis, is $\frac{\pi}{3}+\frac{\pi}{3}=\frac{2 \pi}{3}$ The equation of angle bisector is $$ \begin{array}{ll} \Rightarrow & \mathrm{y}=\mathrm{mx}=\tan \frac{2 \pi}{3} \cdot \mathrm{x} \\ \Rightarrow & \mathrm{y}=-\sqrt{3} \cdot \mathrm{x} \Rightarrow \sqrt{3} \cdot \mathrm{x}+\mathrm{y}=0 \end{array} $$

Hence, the correct answer is option: 'D'.